A simple module

Question

Suppose $R$ is a ring and $M$ is an $R$-module. Prove:

$M$ is simple if and only if there is a left maximal ideal $m$ such that $M\cong R/m$.

Answer 1

You need $\mathfrak{m}$ is a maximal left ideal?

If $M$ is simple, take $0\neq m$ and let $\mathfrak{m}$ be the kernel of the surjection $R\rightarrow M \quad r\mapsto rm$

If $M=R/\mathfrak{m}$, then there are no submodules $\neq 0$ because $\mathfrak{m}$ is maximal.